Discussion Paper Series
Extreme Idealism and Equilibrium in the
Hotelling-Downs Model of Political Competition
David Ronayne
Original version April 2016
Revised Version August 2016
No: 21
Extreme Idealism and Equilibrium in the
Hotelling-Downs Model of Political Competition
David Ronaynea
15 August 2016
Abstract
In the classic Hotelling-Downs model of political competition there is (almost
always) no pure strategy equilibrium with three or more potential strategic candidates where the distribution of voters’ preferred policies are single-peaked. I
study the effect of introducing two idealist candidates who are non-strategic (i.e.,
fixed to their policy platform), to an unlimited number of potential strategic entrants. I present results that hold for a non-degenerate class of cases: (i) For any
equilibrium, it must be that the left-most and right-most candidates (i.e., extremists) are idealists; (ii) Hotelling’s Law fails: in any equilibrium, candidates do not
share their policy platforms, which instead are spread out across the policy space;
(iii) Characterizations for symmetric and asymmetric single-peaked distributions
of voters’ ideal policy preferences. Equilibria where many strategic candidates enter exist only if the distribution of voter preferences is asymmetric. (JEL: C72;
D72)
Keywords: Hotelling; political competition; equilibrium existence; idealism
a
University of Warwick, Coventry CV4 7AL, United Kingdom; [email protected]. I thank Dan
Bernhardt and Peter Buisseret for helpful discussion and comments.
1
1
Introduction
The Hotelling-Downs model of political competition is the workhorse of political scientists
and political economists. It is well known that the unique equilibrium when there are N = 2
strategic candidates is that they both locate at the median voter’s ideal policy. However, in his
seminal work Osborne (1993) shows the negative result that this model, adapted for endogenous entry with N ≥ 3 strategic candidates who maximize their plurality, fails to admit an
equilibrium in pure strategies for all but pathological cases of single-peaked distributions of
voter ideal points.
Equilibrium existence fails generically when N ≥ 3 because there must be two candidates
located at the left-most and right-most policy platforms, which requires the mass of voters to
the left and right of these positions to be equal: an onerous requirement of the distribution of
voter ideal points. These conditions are only met by a degenerate class of distributions including the uniform, which significantly weakened the results of previous studies that employed
such distributions e.g., Cox (1987, Theorem 2). In this article I consider a slight variant of the
classical model, show when this can restore the existence of pure strategy equilibria and offer
a characterization thereof.1
I suppose that in addition to strategic candidates, there are two idealist candidates, who are
fixed to their policy platforms. My first result establishes that for equilibria to generically exist
(within the class of single-peaked distributions), these idealists must be extremists i.e., occupy
left-most and right-most positions. This appeals to the notion that in reality, candidates with
the most extreme positions may often be those unwilling to compromise on their positions.
The classic lesson from Hotelling’s result under N = 2 is that candidates’ incentives to
maximize their vote share will lead them converge on the same (median) platform in equilibrium, a prediction referred to as Hotelling’s Law or the principle of minimum differentiation.
An initial prediction for the nature of equilibrium under larger N was made by Chamberlain
(1933, Appendix C) who conjectured (in relation to firms and candidates) that “they may group
in twos”. The second result I present says that in almost any equilibrium, it must be that there
1
I study pure equilibria in this article and hereon refer to these simply as ‘equilibria’.
2
is exactly one candidate at every policy platform. Combined with the fact that strategic candidates who are willing to enter tie, this implies that their positions are spaced evenly throughout
the distribution of voter preferences. This maximal differentiation of candidate positions in
equilibrium shows that in this setup, Hotelling’s Law fails.
Other researchers have also proposed variations of the canonical model in which a pure
strategy equilibrium obtains. Osborne (1993) defines a dynamic version of the model, and offers results for N = 3 (and partial results for N = 4, 5) showing, among other findings, that
there is always an equilibrium in which N − 2 candidates enter and locate at the median. Xefteris (2016) shows that when one allows for each voter to cast k ≥ 2 votes each instead of just
k = 1, then equilibrium exists for a non-degenerate class of distributions where there are at
least k + 1 candidates at every location. In contrast, I offer results in a plurality voting system
(common to many countries e.g., United States, Canada, India and United Kingdom) where the
number of potential strategic candidates, who may choose whether to enter in equilibrium, is
unlimited i.e., N = ∞.
I characterize the equilibria for a non-degenerate set of single-peaked distributions of voter
ideal points. For symmetric distributions, there is a unique equilibrium in which one strategic candidate enters and wins the election outright when the idealists are not too extreme or
too moderate, relative to the distribution of voter preferences. I then show that equilibria with
multiple strategic entrants exist only if the distribution of voter ideal points is asymmetric, but
that the converse is not true. I then provide a characterization for equilibria under asymmetric
single-peaked distributions. I also give examples of these equilibria for various symmetric and
asymmetric distributions.
2
Model
The model setup is the same as that of the canonical Hotelling-Downs model, but allows for
an endogenous entry, an unlimited number of candidates, and a reasonable objective function
for candidates. The policy space is represented by some interval X ⊆ R. The ideal policies of
voters are spread out along X by an atomless distribution function F which is assumed continuous, guaranteeing it has a density, f . Voters are assumed to be sincere and to have symmetric
3
preferences around their ideal points, meaning that they vote for the candidate positioned closest to that point. There is an unlimited number of strategic candidates (i.e., N = ∞) and two
idealist candidates. Idealist candidates always enter and occupy positions denoted z1 , z2 ∈ X
where z1 < z2 .2 Strategic candidates each have the action set X ∪ {out} i.e., they either enter
and choose a policy platform denoted xi , or they choose not to enter the race. The number of
strategic candidates choosing to enter the race is denoted n and the vector of positions x. Candidates who do not enter are inactive. The utility of a candidate who stays out is normalized
to zero. The functions vi : X n → [0, 1] denote the share of votes obtained by each candidate i
given a vector of positions x.
Strategic candidates maximize their plurality i.e., their margin of victory. Their preferences
are represented by the following utility function:
ui (x) = vi (x) − max{vl (x)}
l6=i
An oft-used objective function for candidates is that of vote maximization. However, vote
maximization is not a reasonable objective function for candidates when N > 2, as it is incompatible with preferences in which winning an election is preferred to losing it (Osborne,
1995, p.280). To illustrate, consider the following example: X = [0, 1], f uniform and position vector xA = (0, 0.5, 0.8) which gives v1 (xA ) = 0.25, v2 (xA ) = 0.4, v3 (xA ) = 0.35
and a victory for candidate 2. Now consider xB = (0, 0.2, 0.8) i.e., candidate 2 moves left,
which gives v1 (xB ) = 0.1, v2 (xB ) = 0.4, v3 (xB ) = 0.5 and a victory for candidate 3. Under
vote-maximization, candidate 2 should be indifferent between xA and xB yet wins the election
under xA and loses under xB . Plurality maximization does not suffer this criticism, saying that
candidates prefer to: (1) stay out rather than enter and lose; (2) win (or tie for the win) than to
lose; and (3) win outright by wider margins.
There are r + 1 ≤ n + 2 occupied positions denoted y0 , . . . , yr indexed without loss of
generality such that y0 < · · · < yr . The midpoints of two adjacent locations is denoted
mj = 21 (yj + yj+1 ). The number of candidates located at yj is denoted kj . The constituency
of a position yj is the share of voters that vote for one of the candidates at yj . The left (right)
2
Although z1 and z2 refer to locations, sometimes I also call the idealists z1 and z2 .
4
constituency of yj denotes the mass of voters voting for a candidate at yj who have ideal points
to the left (right) of yj , denoted Lj , Rj i.e., Lj = F (yj ) − F (mj−1 ) and Rj = F (mj ) − F (yj )
for j = 0, . . . , r but with F (m−1 ) ≡ 0 and F (mr ) ≡ 1.
3
Results
I first present some necessary conditions for an equilibrium to exist generically within the
class of single-peaked densities f . These include the results that idealists must be the extreme
candidates and that Hotelling’s Law fails. Proofs and intermediate Lemmas are found in the
Appendix. I then add sufficient conditions in order to characterize equilibria for symmetric and
asymmetric single-peaked distributions.
Proposition 1 (Extreme idealism). For almost any single-peaked f : y0 = z1 , yr = z2 and
k1 = kr = 1 in equilibrium.
Proposition 1 reveals that the left-most and right-most (extreme) positions must be occupied by idealists for an equilibrium to exist for almost any single-peaked f . With only strategic
candidates, it must be that k0 = kr = 2 and therefore that L0 = R0 and Lr = Rr (Lemma A1,
b and c) which are so restrictive that they preclude equilibrium in all but a degenerate class
of distributions (Osborne, 1993). In contrast, when extreme positions are occupied by candidates who are void of strategic concerns these requirements do not arise. This gives rise to
Proposition 1: in any equilibrium, extremists must be idealists.
Proposition 2 (Hotelling’s Law fails). For almost any single-peaked f , kj = 1 for all j when
n ≥ 2.
Proposition 1 dealt with the extreme locations. Proposition 2 deals with the intermediate
positions and shows that these also cannot hold two strategic candidates in equilibrium, except for very special cases of F . Due to the endogenous entry decision all strategic candidates
who enter, tie in equilibrium (Lemma A1, d). Together with Proposition 2 this implies that a
necessary condition of equilibrium is that the strategic candidates are spaced evenly throughout the distribution of voter preferences. However, Hotelling’s Law stipulates that candidates
5
are incentivized to converge upon shared locations. The separation of candidates’ equilibrium
positions here shows this law can fail, irrespective of the number of strategic entrants.
The results of Propositions 1 and 2 lay the ground work for the equilibrium characterizations. Firstly, Proposition 3 provides the conditions for which an equilibrium exists for a
non-degenerate class of symmetric single-peaked distributions of voter preferences.
Proposition 3 (Symmetric distributions). For almost any symmetric, single-peaked f , there is
a unique equilibrium where n = 1 strategic candidate enters at location y1 , where y1 solves (1):
F (m0 ) = 1 − F (m1 )
(1)
whenever the positions of the idealists (z1 , z2 ) satisfy (2) and (3):
(2) Not too moderate: m0 < F −1
1
3
⇐⇒ m1 > F −1
2
3
(3) Not too extreme: If z1 is closer to the peak of f than z2 , F (y1 ) ≥ 1 − 2F (m0 ).
If z2 is closer to the peak of f than z1 , F (y1 ) ≤ 2F (m0 ).
Except for single-peakedness, the conditions of Proposition 3 deliver equilibrium existence
without other restrictions on the shape of f . Condition (1) is implied by the requirement that the
idealists’ vote-shares must be equal in equilibrium (if not, then due to symmetry the strategic
candidate could profitably deviate by moving slightly towards the idealist with the higher vote
share). Conditions (2)-(3) state that relative to the distribution of voter preferences, the idealists cannot be too moderate or too extreme. They cannot be too moderate because a strategic
candidate must win. They cannot be too extreme else there is room for an entrant to deviate in
and win. I now illustrate the characterization with two examples.
Example 1: Let F be the standard Normal Distribution and the idealists be located at
the 15th and 90th percentiles: (z1 , z2 ) = (F −1 (0.15), F −1 (0.90)) = (−1.04, 1.28). Condition (1) then gives y1 = −0.12. The remaining conditions are also satisfied: (2) becomes
m0 = −0.58 < −0.43 = F −1 31 and the first statement of (3) becomes F (y1 ) = 0.45 ≥
0.44 = 1 − 2F (m0 ). The left panel of Figure 1 shows this equilibrium.
Example 2: Let F be the triangular distribution with the density f (x) = 1 − |x| for
x ∈ [−1, 1] and the idealists be located at the 1st and 77th percentiles: (z1 , z2 ) = (F −1 (0.01),
F −1 (0.77)) = (−0.86, 0.32). Condition (1) then gives y1 = 0.26. The remaining conditions
6
are also satisfied: (2) becomes m0 = −0.30 < −0.18 = F −1
1
3
and the second statement
of (3) becomes F (y1 ) = 0.73 ≥ 0.51 = 1 − 2F (m0 ). The right panel of Figure 1 shows this
equilibrium.
Figure 1: Equilibrium for symmetric single-peaked distributions
f (x)
z1 y1
z1
z2
y1 z2
Left panel: Differentiable f with unbounded support (the standard Normal).
Right panel: Non-differentiable f with bounded support (a triangular distribution).
The shaded area is the constituency of the winning candidate.
A feature of Proposition 3 is that with symmetric single-peaked densities, only one strategic
candidate enters in equilibrium. In Corollary 1, I show that this feature is not special to symmetry per se: it will hold generically in equilibrium for any single-peaked distribution where
the mode (Mo) equals the median (Md).
Corollary 1. For almost any single-peaked f where Mo(f ) = Md(f ), n = 1.
To understand the result, suppose instead that n > 1. I show in the Appendix that exactly
one idealist loses (Lemma A4). Further, there cannot be more than one strategic candidate with
any of their constitution on the same side of the mode as the losing idealist (else candidate closest to the losing idealist could profitably deviate by moving slightly towards the mode). With
n > 1, there would then be at least one strategic candidate with their whole constituency on
the same side of the mode as the idealist who ties for the win. However, for these candidates to
win, there must be more than half the probability density on that side of the mode, contradicting
Mo(f ) = Md(f ).
I now characterize equilibria where n > 1 strategic candidates enter. By Corollary 1 we
know that distributions of voter preferences that support such equilibria are such that Mo(f ) 6=
7
Md(f ), and hence are necessarily asymmetric. Furthermore, the simple fact of whether the
median or the mode of f is greater will play a role in determining equilibria. Proposition 4 provides conditions for an equilibrium to exist for asymmetric single-peaked distributions of voter
preferences where Mo(f ) 6= Md(f ): Equilibria in which multiple strategic candidates enter exist for a non-degenerate set of distributions and idealist positions. Figure 2 gives two examples.
Proposition 4 (Asymmetric distributions). For almost any asymmetric, single-peaked f satisfying (4) - (6) where Mo(f ) 6= Md(f ), there is an equilibrium with n > 1 strategic candidates
where locations and vote-shares are given by Lemma A7.
If Mo(f ) < Md(f )
If Mo(f ) > Md(f )
(4)
f (m0 ) ∈ [f (m1 ), 2f (m1 )]
f (mn ) ∈ [f (mn−1 ), 2f (mn−1 )]
(5)
f (mj ) ≤ 2f (mj+1 ) j = 1, . . . , n
f (mj−1 ) ≤ 2f (mj ) j = 1, . . . , n
(6)
f (m0 ) ≤ max{f (y1 ), f (z1 )}
f (mn ) ≤ max{f (yn ), f (z2 )}
Compared to the symmetric case, there are more equilibrium conditions when F is asymmetric and n > 1. Lemma A7 provides conditions (A6) and (A9) which are analogous to condition (2) of Proposition 3, saying that the losing idealist must be extreme enough to lose. The
Lemma also provides the exact equilibrium location of strategic candidates (conditions A4, A5,
A7, A8) which as Proposition 2 revealed, are spaced out evenly through the distribution of voter
ideal points. Specifically, the locations of the idealist candidates pin down the vote share, s∗ ,
enjoyed by each of the strategic candidates in equilibrium. The strategic candidates’ locations
are then determined by the following ‘spacing procedure’ (detailed in precisely in Lemma A5):
To illustrate, suppose that z1 ties for the win (which is the case if Mo(f ) < Md(f )); then place
the first strategic candidate at y1 , such that z1 has a vote share of s∗ ; then place the second
strategic candidate at y2 , such that the candidate at y1 has a vote share of s∗ , and so on; the
losing idealist, in this case z2 , will then be left with the residual vote share of 1 − s(n + 1).
For asymmetric distributions, there are also conditions concerning the shape of f , given by
(4)-(6). The requirements of (4) and (5) that f (mj ) ≤ 2f (mj+1 ) for j = 0, . . . , n are driven by
the fact that strategic candidates are plurality maximizers. To see this, consider the candidate
at y2 in the upper panel of Figure 2, and a deviation slightly to the left. This reduces the vote
8
share of the candidate at y1 , but raises that of z2 . The marginal gain in plurality is f (m1 ), but
the marginal loss is 2f (m2 ): f (m2 ) for the loss in vote share and another f (m2 ) for the gain
in vote share of z2 . Therefore, if f (m1 ) > 2f (m2 ) there would be such a deviation. Also, (6)
requires that the density of the midpoint between the losing extremist and their neighboring
strategic candidate not be higher than the density of either of those candidate’s locations. The
condition precludes the possibility that there could be a profitable deviation for an inactive candidate to enter and locate in such a way that they win the election with the peak of f in their
constituency. All conditions are met by the examples in Figure 2 which are therefore equilibria
with asymmetric, single-peaked distributions of voter preferences.
Figure 2:
Equilibrium for asymmetric single-peaked distributions with n > 1 strategic candidates
f (x)
z1
y1 y2
z2
f (x)
z1
y1
y2 y3
y4 y5 y6 y7 z2
Top panel: n = 2; Log-Normal distribution ln N(0, 0.5); idealists at 1st and 83rd percentiles.
Bottom panel: n = 7; Linear distribution; idealists at the 7th and 100th percentiles.
The shaded areas are the constituencies of the winning candidates.
More generally, the results of this article do not change qualitatively if other extreme idealist
candidates are added appropriately. For example, more idealists can be added at any positions
9
to the left (right) of z1 (z2 ) when z1 (z2 ) loses.3 Rather than having two idealists, one can
contemplate potentially many fringe candidates on the extremes of the political spectrum.
I established a simple relationship between the mode and median of f as a determinant of
the number of candidates entering in equilibrium. Such a measure cannot of course hope to
capture every way in which distributions can be asymmetric, but nevertheless acts as a succinct
predictive measure in plurality voting systems with idealist candidates. With multiple strategic
candidates in such an election, it must be that the distribution of voter preferences is such that
the mode and median are distinct. Equally, if the distribution of voter preferences is symmetric
and there are fringe idealists on each extreme, one strategic candidate will run and win.
References
Chamberlain, Edward, 1933, The Theory of Monopolistic Competition (Harvard University
Press, Cambridge, MA).
Cox, Gary W, 1987, Electoral equilibrium under alternative voting institutions, American
Journal of Political Science 31(1), 82–108.
Downs, Anthony, 1957, An Economic Theory of Democracy (Harper & Row, New York).
Hotelling, Harold, 1929, Stability in competition, Economic Journal 3, 41–57.
Osborne, Martin J, 1993, Candidate positioning and entry in a political competition, Games
and Economic Behavior 5(1), 133–151.
1995, Spatial models of political competition under plurality rule: A survey of some
explanations of the number of candidates and the positions they take, The Canadian Journal
of Economics 28(2), 261–301.
Xefteris, Dimitrios, 2016, Stability in electoral competition: A case for multiple votes, Journal
of Economic Theory, forthcoming.
3
Idealists can also be added to the left (right) of z1 (z2 ) when z1 (z2 ) wins, but it requires a minor reworking
to the expressions in the results presented here to reflect the fact the strategic candidate’s equilibrium locations
need to be shifted to ensure z1 (z2 ) still ties for the win.
10
Appendix
Lemma A1. When at least one strategic type is at yj :
(a) kj ≤ 2.
(b) kj = 2 for j = 0, r.
(c) If kj = 2, Lj = Rj .
(d) All strategic candidates who enter, tie and win.
Proof: See Cox (1987) and Osborne (1993) Lemma 1 who prove these when all candidates are
strategic. If none of the candidates at the positions in question are strategic, then the proofs do
not apply, hence the qualified version of their Lemmas here.
Lemma A2. For almost any distribution F , not all candidates tie.
Proof: Suppose not. Firstly, consider the case where there are two candidates at an extreme
location and without loss of generality, suppose this is on the left i.e., k0 = 2. By Lemma A1
2
3
1
and m0 = F −1 n+2
. If k1 = 2, then y1 = F −1 n+2
which implies
(c), y0 = F −1 n+2
1
3
2
F −1 n+2
+F −1 n+2
= 2F −1 n+2
, which is not satisfied for almost any distribution. Continuing similarly, one shows that generically, kj = 1 for all j > 1 (see the proof of Lemma 2 in
Osborne, 1993 which I have presented an adapted version of, up to this point). It must be there
j+2
fore that r = n and yr = z2 . For all candidates to tie, mj = F −1 n+2
for j = 0, . . . , n − 1.
P
j −1 j+2
Solving recursively yields y0 = (−1)n z2 +2 n−1
. However, we also required
j=0 (−1) F
n+2
1
. These two expressions are not satisfied for almost any distribution.
y0 = F −1 n+2
Now consider the case where there is one candidate at each extreme location k0 = kr = 1,
which by Lemma A1 implies y0 = z1 and yr = z2 . For all to tie, F (mj ) = F (mj−1 ) + sj
k
j
and F (m−1 ) ≡ 0. Solving recursively yields z1 =
for all j = 0, . . . , r − 1 where sj = n+2
P
P
j −1
(−1)r z2 + 2 r−1
(Sj ), where Sj = ji=1 si which is not true for almost any F .
j=0 (−1) F
Proposition 1 (Extreme idealism). For almost any single-peaked f : y0 = z1 , yr = z2 and
k1 = kr = 1 in equilibrium.
11
Proof: Suppose not. Either k0 = 2 or kr = 2 by Lemma A1 (b). Without loss of generality say
k0 = 2, which implies L0 = R0 by Lemma A1 (c). Denote the equilibrium vote share of the
winning candidates by s.
If n = 1 this imposes F (z1 ) = F ( 12 (z1 + z2 )) − F (z1 ), which is not true for almost any F .
If n = 2, s ≥ 14 . If s = 41 , all candidates tie, which is ruled out by Lemma A2. If s > 41 , then by
Lemma A1 (d), z2 is the sole loser. It must be that the strategic candidate is located at y1 < z2 :
if they were located at z2 , then they would tie with z2 ; if they were located right of z2 , they could
profitably deviate slightly to the left. If f (m0 ) > f (m1 ), then the candidate at y1 can profitably
deviate by moving slightly to the left (they increase their share, and decrease the winning candidates’ shares). If f (m0 ) ≤ f (m1 ), R0 < L1 because f is single-peaked. But L0 = R0 = s,
hence the candidate at y1 must get strictly more than s votes and wins outright, a contradiction.
For n ≥ 3 strategic candidates, y0 = F −1 (s) and m0 = F −1 (2s). If there is a strategic candidate at y1 and k1 = 2, then y1 = F −1 (3s) which implies 21 (F −1 (s) + F −1 (3s)) = F −1 (2s),
which is true for almost no distribution F . Hence k1 = 1 and m1 = F −1 (3s). Similarly, if there
is a strategic candidate at y2 , k2 = 1 for almost any F , and so on. Denote yi as the left-most
position after y0 where there is an idealist. What I have shown so far is that for almost any F ,
ki = 1. Notice that by Lemma A1 (d) and Lemma A2, z2 must lose for almost all F . Now I
consider two cases, both of which end in a contradiction.
(i) If there are no strategic candidates to the right of yi , then for the single-peaked density
f : if f (mi−2 ) ≤ f (mi−1 ), then L1 > s because R0 = s, which contradicts Lemma A1 (d); if
f (mi−2 ) > f (mi−1 ), then the candidate at yi has a profitable deviation slightly to the left (by
increasing their own vote share and decreasing that of the winning candidates).
(ii) If there is a strategic candidate to the right of yi , let the right-most such candidate be at
yj . If f (mi−1 ) ≤ f (mi ), L1 > s. If f (mi−1 ) > f (mi ), I consider two sub-cases: If j = r, then
kr = 2 and Rr = Lr = s < Rr−1 . If j < r, then j = r − 1 and there is a lone idealist at yr ,
in which case yj can deviate profitably by moving slightly to the left (by increasing their own
vote share and decreasing that of the winning candidates).
Lemma A3. For almost any distribution F , kj = 1 for all j when n = 2.
12
Proof: Suppose not. Then by Proposition 1 and Lemma A1 (c), k1 = 2 and L1 = R1 . If z1 gets
a strictly lower (higher) vote share than z2 , an entrant can locate slightly to the right (left) of the
strategic candidates at y1 and win outright. Thus all candidates tie, contradicting Lemma A2.
Lemma A4. For almost any single-peaked f , exactly one idealist must tie with the strategic
candidates when n ≥ 2.
Proof: First, I show that both idealists cannot lose. Suppose they do and consider first n = 2.
By Lemma A3, k1 = k2 = 1. If f (m0 ) < f (m1 ), the candidate at y1 can move slightly to
the right, increasing their vote-share and decreasing that of the other strategic candidate; if
f (m0 ) ≥ f (m1 ) then because f is single-peaked, the maximizer of f must lie to the left of
m1 , which implies f (m1 ) > f (m2 ) and hence that the candidate at y2 can profitably deviate by
moving slightly to the left, increasing their vote-share and decreasing that of the other strategic
candidate.
Now consider n ≥ 3. Denote the equilibrium vote share of strategic candidates as s. If y2
is to the left (weakly) of the maximizer of f , then k1 = 1 because if k1 = 2, s = R1 < L2 ,
contradicting Lemma A1 (d). As k1 = 1 and z1 loses, the candidate at y1 can profitably deviate
slightly to the right. Now consider the case where y2 is to the right of the maximizer. There can
be no more strategic candidates to the right of y2 . If there were, then kj = 1, j > 2 because if
kj = 2 for one such j, then Rj−1 > Lj = s. Note now that the candidate at yr−1 has a profitable
deviation to the left because z2 loses. Next I show that it must be that k1 = k2 = 2 and hence
that n = 4. If k2 = 1 and f (m1 ) > f (m2 ), the candidate at y2 can profitably deviate to the left;
if k2 = 1 and f (m1 ) ≤ f (m2 ), k1 = 1 (else s = R1 < L2 ) and the candidate at y1 can profitably
deviate right. Hence k2 = 2. If k1 = 1 and f (m0 ) < f (m1 ), the candidate at y1 can profitably
deviate right; if k1 = 1 and f (m0 ) ≥ f (m1 ) then f (y1 ) > f (m1 ) implying R1 > L2 = s as
k2 = 2. As k1 = k2 = 2, by Lemma A1 (c) and (d), L1 = R1 = L2 = R2 . But with only two
free variables (y1 and y2 ) these three conditions will not be satisfied for almost any F .
Hence, for almost all single-peaked distributions at least one idealist must tie, but by
Lemma A2, exactly one idealist must tie.
13
Lemma A5. For almost any single-peaked f , kj = 1 for all j when n ≥ 3.
Proof: By Proposition 1, y0 = z1 and yr = z2 while by Lemma A1 (d) all strategic entrants tie
for the win. This implies F (z1 ) <
1
n+2
and F (z2 ) >
n+1
n+2
in any equilibrium. By Lemma A4,
exactly one idealist ties with the strategic types and without loss of generality, let this be z1 .
Now consider the following spacing procedure which spaces candidate locations throughout
the distribution F for some arbitrary number of candidates n, where k0 = kr = 2, kj = 1, 2 for
j = 1, . . . , r − 1 and strategic types tie with the idealist z1 .
Spacing Procedure:
1
.
1. Choose y1 such that s ≡ F (m0 ) ∈ F (z1 ), n+2
2. Place the remaining r − 2 candidate locations at yj for j = 2, ..., r − 1 in turn, such that
F (mj−1 ) = F (mj−2 ) + kj−1 s.
3. Observe whether 21 (yr−1 + z2 ) = mr−1 . If yes, stop and denote s as s∗ ; if mr−1 < (>)
1
(y
2 r−1
+ z2 ) return to step 1 and choose a higher (lower) value of s.
Iterating on this procedure, the value of s will converge to s∗ . As F is continuous, s∗ exists,
and as F is strictly increasing, s∗ is unique. An example result of the procedure is illustrated below in Figure A1. The points y1 , . . . , yr−1 associated with s∗ pin-down the necessary locations
of the strategic candidates in equilibrium.4
4
Notice that although s∗ is necessarily the equilibrium share of the vote for the winning candidates, this proce-
dure is not sufficient to define an equilibrium as for example, it may not be that yj > yj−1 for all j = 1, ..., r − 1.
14
Figure A1: An example result of the spacing procedure
5s∗ = F (m3 )
F (x)
4s∗ = F (m2 )
R2
2s∗ = F (m1 )
L2
∗
s = F (m0 )
z1
y1 y2
y3 z2
The example shown has n = 4 and r = 4 where ki = 1 for all i except k2 = 2. F
is the standard Normal distribution and z1 = F −1 (0.10), z2 = F −1 (0.98). Solving
the procedure yields s∗ = 0.19 (2 d.p.) with candidate positions as shown.
It is now straightforward to see that for almost any distribution F , ki = 1 for all i. Suppose
instead that kj = 2 for some j = 2, ..., r. By Lemma A1 (c) we must have that Lj = Rj .
However, as is illustrated in Figure A1 for the example of j = 2, this extra condition will not
be satisfied for all except very particular distributions.
Proposition 2 (Hotelling’s Law fails). For almost any single-peaked f , kj = 1 for all j when
n ≥ 2.
Proof: Immediate from Lemmas A3 and A5.
Lemma A6. For any symmetric, single-peaked f , when there is n = 1 strategic entrant, the
idealists’ vote shares are equal.
Proof: Suppose not. Without loss of generality, suppose that the idealist z1 has a higher vote
share than z2 which implies that f (m0 ) > f (m1 ). The strategic candidate at y1 can move
slightly to the left, simultaneously increasing their own vote share and reducing the vote share
of z2 , giving strictly higher utility.
15
Proposition 3 (Symmetric distributions). For almost any symmetric, single-peaked f , there is
a unique equilibrium where n = 1 strategic candidate enters at location y1 , where y1 solves (1):
F (m0 ) = 1 − F (m1 )
(1)
whenever the positions of the idealists (z1 , z2 ) satisfy (2) and (3):
(2) Not too moderate: m0 < F −1
1
3
⇐⇒ m1 > F −1
2
3
(3) Not too extreme: If z1 is closer to the peak of f than z2 , F (y1 ) ≥ 1 − 2F (m0 ).
If z2 is closer to the peak of f than z1 , F (y1 ) ≤ 2F (m0 ).
Proof: Firstly I show that n = 1 in equilibrium. Suppose instead n > 1. By Lemmas 1, A3 and
A5, for almost all single-peaked f , the strategic candidates occupy the non-extreme locations
and kj = 1 for all j. As f is symmetric, there must be at least one strategic candidate on either
side of the maximizer of f , else Lemma A1 (d) is violated. I now show this implies that both
idealists tie with the strategic candidates. Suppose not and without loss of generality that z1
loses. As f is symmetric, this implies f (m0 ) < f (m1 ) (if not, z1 gets at least as many votes
as the candidate at y2 ). The candidate at y1 then can profitably deviate slightly to the right. But
by Lemma A2 for almost all distributions F , not all candidates can tie.
I now characterize the equilibrium. By Lemma A6, the idealists’ vote shares must be equal,
meaning that the strategic candidate’s position y1 must solve (1). To be an equilibrium, the
strategic candidate must win, which implies F (m1 ) − F (m0 ) > 31 . Using (1), this becomes (2).
In equilibrium, the strategic candidate must not want to deviate to the left of z1 or the right of
z2 . Note that (2) implies that z1 < F −1 ( 13 ) and z2 > F −1 ( 23 ). As the strategic candidate gets at
least
1
3
of the vote share in order to win, there is no such profitable deviation. The strategic can-
didate would also lose if they deviated to an idealist’s location as the other idealist would win
outright. Finally, the strategic candidate does not have incentive to deviate to another location
in (z1 , z2 ): Without loss of generality, consider such a deviation to the left. By Lemma A6 this
increases z2 ’s vote share (and z2 now beats rather than ties with z1 ). However, as f symmetric,
this deviation also decreases the strategic candidate’s vote share and hence also their plurality.
In equilibrium, inactive strategic candidates must not wish to enter. Notice that an inactive
candidate could only profitably locate in (z1 , z2 ). Assume first that z1 is closer to the maximizer
16
of f than z2 , so that y1 is to the left of the maximizer. Notice that the payoff of the entrant is increasing as their location approaches y1 from the right. Hence, entry is not profitable if the right
constituency of y1 is less than the vote share of the idealists F (m1 ) − F (y1 ) ≤ F (m0 ) which
gives (3). Similarly, the case of z2 being closer to the peak gives the other expression in (3).
Corollary 1. For almost any single-peaked f where Mo(f ) = Md(f ), n = 1.
Proof: Suppose instead n > 1. By Lemma A4 exactly one idealist loses and without loss of
generality assume this is z2 . This implies that f (mr−2 ) ≤ f (mr−1 ) else the candidate at yr−1
deviates left. This implies that mr−2 is strictly to the left of the maximizer of f . For the candidate at yr−2 and z1 to tie (along with any number of others on the left of the maximizer),
there must be strictly more than half the density to the left of the maximizer, contradicting
Mo(f ) = Md(f ).
Lemma A7. For almost any single-peaked f , when n ≥ 2, strategic candidates and one idealist tie for the win with vote share s∗ , where:
If Mo(f ) < Md(f ), then s∗ solves (A4), locations are given by (A5) and the left extremist loses (A6);
(A4)
(A5)
n+1
z1 = (−1)
z2 − 2
n+1
X
(−1)n+i F −1 (1 − is)
i=1
n+1−j
X
yj = (−1)n+1−j z2 − 2
(−1)n−j+i F −1 (1 − is∗ ), s.t. z1 < yj < yj+1 , j = 1, ..., n
i=1
(A6)
z1 < 2F −1 (s∗ ) − y1 .
If Mo(f ) > Md(f ), s∗ solves (A7), locations are given by (A8) and the right extremist loses (A9);
(A7)
(A8)
z1 = (−1)n+1 z2 + 2
yj = (−1)j z1 + 2
n+1
X
(−1)n+i F −1 (is)
i=1
j
X
(−1)j+i F −1 (is∗ ), s.t. z1 < yj < yj+1 , j = 1, ..., n
i=1
(A9)
z2 > 2F −1 (1 − s∗ ) − yn .
Proof: I first show that if Mo(f ) < Md(f ) and n > 1, z1 loses: If not, by Lemma A4 z2 loses
and one can then then follow the proof of Corollary 1 to show that there must be strictly more
17
than half the density to the left of the maximizer, contradicting Mo(f ) < Md(f ). Given z1
loses, z2 must tie with the strategic candidates by Lemma A4 and kj = 1 for all j by Lemmas
A3 and A5. This implies that r = n + 1 and that F (mj ) = F (mj−1 ) + s for j = 1, . . . , n + 1
where s is the equilibrium vote share and F (mn+1 ) ≡ 1. Solving recursively yields (A4) which
the equilibrium s solves, giving equilibrium locations as (A5) where (A6) is the requirement
for z1 to lose: F (m0 ) < s∗ . Similarly, one finds (A7)-(A9) in the case of Mo(f ) > Md(f ).
Proposition 4 (Asymmetric distributions). For almost any asymmetric, single-peaked f satisfying (4) - (6) where Mo(f ) 6= Md(f ), there is an equilibrium with n > 1 strategic candidates
where locations and vote-shares are given by Lemma A7.
If Mo(f ) < Md(f )
If Mo(f ) > Md(f )
(4)
f (m0 ) ∈ [f (m1 ), 2f (m1 )]
f (mn ) ∈ [f (mn−1 ), 2f (mn−1 )]
(5)
f (mj ) ≤ 2f (mj+1 ) j = 1, . . . , n
f (mj−1 ) ≤ 2f (mj ) j = 1, . . . , n
(6)
f (m0 ) ≤ max{f (y1 ), f (z1 )}
f (mn ) ≤ max{f (yn ), f (z2 )}
Proof: I show that conditions (4) - (6) are sufficient for an equilibrium by considering all possible deviations in the case of Mo(f ) < Md(f ); those for Mo(f ) > Md(f ) follow similarly.
Consider deviations of the candidate at y1 within (z1 , y2 ) (the candidate at y1 is the only
strategic candidate who could have a constituency boundary to the left of the maximizer of f )
(i) to the left: the candidate at y2 then becomes the candidate with the highest vote-share of
all other candidates, hence if f (m0 ) ≤ 2f (m1 ) there is no profitable deviation within (z1 , y1 );
(ii) to the right: for a small move, z1 remains a loser and the candidate at y2 becomes a loser.
It must be that f (m0 ) ≥ f (m1 ) else the candidate at y1 could profit from such a move. This
implies that any deviation within (y1 , y2 ) reduces this candidate’s vote share, hence there is no
such profitable deviation. This gives (4).
Next consider deviations for the candidate at yj , j > 1 within (yj−1 , yj+1 ) (i) to the left:
their vote share would increase, but so will that of the candidate at yj+1 who then becomes the
candidate with the highest share of all the others, but the plurality of the deviating candidate
decreases if f (mj−1 ) ≤ 2f (mj ) which gives (5); (ii) to the right: their own vote share would
18
decrease while increasing that of the candidate at yj−1 .
Next consider an inactive candidate entering (i) at an occupied location: this is not profitable as it results in an outright loss; (ii) left of z1 or right of z2 : this results in an outright loss;
(iii) between two strategic candidates yj and yj+1 , j > 1: such an interval does not contain
the maximizer of f , hence the optimal such deviation is as close as possible to the candidate
whose position is has higher density, yj . But this cannot be profitable because the maximum
vote share is bounded from above by max{Lj , Rj } < s∗ ; (iv) between z1 and y1 , which contains the maximizer of f : under (6), the optimal such deviation is to locate arbitrarily close
to z1 or y1 (whichever has the higher density), but as in case (iii) this is unprofitable because
max{R0 , L1 } < s∗ .
Finally, for deviations of the candidate at yj to locations outside the interval (yj−1 , yj+1 ),
j = 1, . . . , n, it suffices to follow the steps above relating to an inactive candidate.
19